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NCERT solutions for Class 12 Physics Chapter 3 Current Electricity

NCERT solutions for Class 12 Physics Chapter 3 Current Electricity

Edited By Vishal kumar | Updated on Sep 08, 2023 03:02 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Physics Chapter 3 – Access and Download PDF for Free

NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity - Welcome to the updated current electricity class 12 exercise solutions for one of the easy and intriguing chapters in the Class 12 syllabus. On this page, you will discover comprehensive NCERT solutions spanning from question 3.1 to 3.23. Questions 3.1 to 3.14 are part of the exercise section, while the rest are from the additional exercise. These class 12 physics chapter 3 ncert solutions have been meticulously crafted by subject experts, aligning with the latest syllabus and offering detailed explanations.

Physics Class 12 chapter 3 NCERT solutions are related to Ohm’s law and Kirchoff's law which will help in solving circuit related problems in exams. The NCERT Solutions for Class 12 Physics Chapter 3 also explains problems related to heat developed in a resistor. The basics of these chapters are covered in NCERT of high school classes also.

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Current is the flow of electric charge through a given area. If we want to utilize the charge flow, we must have a closed path. NCERT Solutions For Class 12 Physics chapter 3 explains questions about the factors affecting the current flow, how current flows in a conductor, basic laws related to current electricity and circuit analysis and a few measuring devices. NCERT Solutions for Class 12 contain the answer to each question given in NCERT Class 12 Physics book.

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NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

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NCERT solutions for class 12 physics chapter 3 Current Electricity: Exercise Question

3.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 \Omega , what is the maximum current that can be drawn from the battery?

Answer :

Given, the emf of battery, E = 12 V

Internal resistance of battery, r = 0.4 Ohm

Let I be the maximum current drawn from the battery.

We know, according to Ohm's law

E = Ir

I = E/r = 12/0.4 =30 A

Hence the maximum current drawn from the battery is 30 A.

3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer:

Given, The emf of the battery, E = 10 V

The internal resistance of the battery, r = 3 Ohm

Current in the circuit, I = 0.5 A

Let R be the resistance of the resistor.

Therefore, according to Ohm's law:

E = IR' = I(R + r)

10 = 0.5(R + 3)

R = 1 Ohm

Also,

V = IR (Across the resistor)

= 0.5 x 17 = 8.5 V

Hence, terminal voltage across the resistor = 8.5 V

3.3 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?

Answer:

We know that when resistors are combined in series, the effective resistance is the sum of that resistance.

Hence, total resistance of the three resistance combination = 1 + 2 + 3 = 6 Ohm

3.3 (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Answer:

Since the resistances are in series, the current through each one of them will be equal to the current through the circuit but voltage/ potential drop will be different.

Total resistance, R = 6 Ω

Emf, V = 12 V

According to Ohm's law:

V = IR

\implies 12 = I x 6

\implies I = 12/6 = 2 A

Now, using the same relation, voltage through resistors:

1 Ω : V(1) = 2 x 1 = 2V

2 Ω : V(2) = 2 x 2 = 4V

3 Ω : V(3) = 2 x 3 = 6V

(Note: V(1) + V(2) + V(3) = 2 + 4 + 6 = 12 V )

3.4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?

Answer:

We know that when resistances are in parallel combination, the total resistance R is given by:

\frac{1}{R} = \sum \frac{1}{R_{i}}

Therefore, total resistance of the given three resistances in parallel combination is

\frac{1}{R} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}

\implies \frac{1}{R} = \frac{1}{2}+\frac{1}{4}+\frac{1}{5} = \frac{10+5+4}{20}=\frac{19}{20}

\implies R= \frac{20}{19} = 1.05 \Omega

Hence, the total resistance is 1.05 Ω

3.4 (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Answer:

Since the resistances are in parallel, the voltage across each one of them will be equal.

Emf, V = 20 V

According to Ohm's law:

V = IR \implies I = V/R

Therefore, current across each one of them is:

2 Ω : I = 20/2 = 10 A

4 Ω : I = 20/4 = 5 A

5 Ω : I = 20/5 = 4 A

3.5 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.7 x 10-4 oC-1

Answer:

Given,

temperature coefficient of filament,

α = 1.7 x 10-4 oC-1

T_1= 27\degree C ; R_1 = 100 \Omega

Let T2 be the temperature of element, R2=117 Ω

(Positive alpha means that the resistance increases with temperature. Hence we can deduce that T2 will be greater than T1 )

We know,

R_2 = R_1[1 + \alpha \Delta T]

\implies 117 = 100[1 +(1.70\times 10^{-4})(T_2 - 27) ]

\\ \Rightarrow T_2-27=\frac{117-100}{1.7\times10^{-4}}\\\Rightarrow T_2-27=1000\\\Rightarrow T_2=1027^\circ C

Hence, the temperature of the element is 1027 °C.

3.7 A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

Answer:

Given,

\dpi{100} T_1= 27.5 \degree C ; R_1 = 2.1\Omega

T_2= 100\degree C ; R_2= 2.7\Omega

We know,

R_2 = R_1[1 + \alpha \Delta T]

2.7 = 2.1[1 + \alpha (100 - 27.5) ]

\alpha = (2.7 - 2.1) / 2.1(100 -27.5)

\alpha = 0.0039 \degree C^{-1}

Hence, the temperature coefficient of silver wire is 0.0039 \degree C^{-1}

3.8 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70\times 10^{-4}\;\degree C^{-1} .

Answer:

For the given voltage, the two values of current will correspond to two different values of resistance which will correspond to two different temperature.

V = 230 V

I_1= 3.2 A and I_2 = 2.8 A

Using Ohm's law:

R_1 = 230/3.2 = 71.87\Omega

and

R_2= 230/2.8 = 82.14 \Omega

T_1= 27\degree C

Let T_2 be the steady temperature of the heating element.

We know,

R_2 = R_1[1 + \alpha \Delta T]

That is

230/2.8 = 230/3.2[1 + ( 1.70\times 10^{-4} ) ( T 2 - 27) ]

T 2 = (840.5 + 27) °C

Hence, steady temperature of the element is 867.5 °C .

3.9 Determine the current in each branch of the network shown in Fig. 3.30:

1643115241787 Answer:

Let current in the circuit is distributed like

1643115269822 where I1, I2, and I3 are the different current through shown branches.

Now, applying KVL in Loop

10-I10-I_25-(I_2+I_3)10=0

Also, we have I=I_1+I_2

so putting it in kvl equation

10-(I_1+I_2)10-I_25-(I_2+I_3)10=0

10-10I_1-10I_2-5I_2-10I_2-10I_3=0

10-10I_1-25I_2-10I_3=0 .................................(1)

Now let's apply kvl in the loop involving I1 I2 AND I3

5I_2-10I_1-5I_3=0 .................................(2)

now, the third equation of KVL

5I_3-5(I_1-I_3)+10(I_2+I_3)=0

-5I_1+10I_2+20I_3=0 ..............................(3)

Now we have 3 equation and 3 variable, on solving we get

I_1= \frac{4}{17}A

I_2= \frac{6}{17}A

I_3= \frac{-2}{17}A

Now the total current

I=I_1+I_2=\frac{4}{17}+\frac{6}{17}=\frac{10}{17}

3.10 (a) In a meter bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is Determine the current in each branch of the network shown in Fig. 3.30:of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

fig327

Answer:

Balance point from the end A, l 1 = 39.5 cm

Resistance of Y = 12.5 Ω

We know, for a meter bridge, balance condition is:

\frac{X}{Y} = \frac{l_1}{l_2} = \frac{l_1}{100 - l_1}

\implies X = \frac{39.5}{100 - 39.5}\times12.5 = 8.2\Omega

The connections between resistors in a Wheatstone or meter bridge made of thick copper strips to minimise the resistance of the connection which is not accounted for in the bridge formula.

3.10 (c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

Answer:

if the galvanometer and cell are interchanged the galvanometer will show no current and hence no deflection.

3.11 A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Answer:

Emf of battery, E = 8 V

Internal resistance of battery, r = 0.5 Ω

Supply Voltage, V = 120 V

The resistance of the resistor, R = 15.5 Ω

Let V' be the effective voltage in the circuit.

Now, V' = V - E

V' = 120 - 8 = 112 V

Now, current flowing in the circuit is:

I = V' / (R + r)

\implies I = \frac{112}{15.5 + 0.5} = 7 A

Now, using Ohm ’s Law:

Voltage across resistor R is v = IR

V = 7 x 15.5 = 108.5 V

Now, the voltage supplied, V = Terminal voltage of battery + V

\therefore Terminal voltage of battery = 120 -108.5 = 11.5 V

The purpose of having a series resistor is to limit the current drawn from the supply.

3.13 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5\times 10^{28}m^{-3} . How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0\times 10^{-6} m^2 and it is carrying a current of 3.0 A.

Answer:

We know,

I = neAv_d

vd :drift Velocity = length of wire(l) / time taken to cover

I = neA \frac{l}{t}

by substituting the given values

\implies t = 2.7 x 104 s

Therefore, the time required by an electron to drift from one end of a wire to its other end is 2.7\times 10^4 s.


NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity Additional Exercises

3.14 The earth’s surface has a negative surface charge density of 10 -9 C m -2 . The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A cover the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10 6 m.)

Answer:

Given, the surface charge density of earth

\rho = 10^{-9}C m^{-2}

Current over the entire globe = 1800 A

Radius of earth, r = 6.37 x 106 m

\therefore The surface area of earth A = 4\pi r^2

= 4\pi (6.37\times10^6 )^2 = 5.09 \times 10^{14} m^2

Now, charge on the earth surface,

q=\rho\times A

Therefore,

1643115342596

Let the time taken to neutralize earth surface be t

\therefore Current I = q / t

t = 282.78 s.

Therefore, time take to neutralize the Earth's surface is 282.78 s

3.15 (a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

Answer:

Given,

There are 6 secondary cells.

Emf of each cell, E = 2 V (In series)

The internal resistance of each cell, r = 0.015 Ω (In series)

And the resistance of the resistor, R = 8.5 Ω

Let I be the current drawn in the circuit.

I = \frac{nE}{R + nr}

\implies I = \frac{6(2)}{8.5 + 6(0.015)} = \frac{12}{8.59}

\implies I = 1.4 A

Hence current drawn from the supply is 1.4 A

Therefore, terminal voltage, V = IR = 1.4 x 8.5 = 11.9 V

3.15 (b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Answer:

Given,

Emf , E = 1.9 V

Internal resistance, r =380 Ω

The maximum current that can drawn is I = E/r = 1.9/380 = 0.005 A

The motor requires a large value of current to start and hence this cell cannot be used for a motor of a car.

3.16 Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. ( \rho_{Al} = 2.63\times 10^{-8} \Omega m . \rho_{Cul} = 1.73\times 10^{-8} \Omega m relative density of Al = 2.7, of Cu = 8.9.)

Answer:

We know,

R = Ρ l / A

The wires have the same resistance and also are of the same length.

Hence,

\frac{\rho_{Al}}{A_{Al}} = \frac{\rho_{Cu}}{A_{Cu}}

\implies \frac{A_{Al}}{A_{Cu}} = \frac{\rho_{Al}}{\rho_{Cu}} = \frac{2.63}{1.73}

Now, mass = Density x Volume = Density x Area x length

Taking the ratio of their masses for the same length

\implies \frac{m_{Al}}{m_{Cu}} = \frac{d_{AL}A_{Al}}{d_{Cu}A_{Cu}} = \frac{2.7\times2.63}{8.9\times1.73} < 1

Hence, m_{Al} < m_{Cu}

Therefore, for the same resistance and length, the aluminium wire is lighter.

Since aluminium wire is lighter, it is used as power cables.

3.17 What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Current

A

Voltage

V

Current

A

Voltage

V

0.2
3.94
3.0
59.2
0.4
7.87
4.0
78.8
0.6
11.8
5.0
98.6
0.8
15.7
6.0
118.5
1.0
19.7
7.0
138.2
2.0
39.4
8.0
158.0

Answer:

The ratio of Voltage to current for the various values comes out to be nearly constant which is around 19.7.

Hence the resistor made of alloy manganin follows Ohm's law.

Answer the following questions:

3.18 (a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?

Answer:

The current flowing through the conductor is constant for a steady current flow.

Also, current density, electric field, and drift speed are inversely proportional to the area of cross-section. Hence, not constant.

Answer the following questions:

3.18 (b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.

Answer:

No. Ohm’s law is not universally applicable for all conducting elements.

A semiconductor diode is such an example.

Answer the following questions

3.18 (c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?

Answer:

Ohm's law states that: V = I x R

Hence for a low voltage V, resistance R must be very low for a high value of current.

Answer the following questions:

3.18 (d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

Answer:

A very high internal resistance is required for a high tension supply to limit the current drawn for safety purposes.

3.19 Choose the correct alternative:

(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of ( 10^{22} / 10^{23} ).

Answer:

(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have much lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of 1022 .

3.20 (a)

(i) Given n resistors each of resistance R , how will you combine them to get the (i) maximum effective resistance?

Answer:

To get maximum effective resistance, combine them in series. The effective resistance will be nR.

3.20 (a)

(ii) Given n resistors each of resistance R, how will you combine them to get the (ii) minimum effective resistance?

Answer:

To get minimum effective resistance, combine them in parallel. The effective resistance will be R/n.

3.20 (a)

(iii) What is the ratio of the maximum to minimum resistance?

Answer:

The ratio is nR/(R/n) = n^2

3.20 (b)

(i) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them
to get an equivalent resistance of (i) \frac{11}{3}\Omega

Answer:

We have, equivalent resistance = 11/3

Let's break this algebraically so that we can represent it in terms of 1, 2 and 3

\frac{11}{3}=\frac{9+2}{3}=3+\frac{2}{3}=3+\frac{1*2}{1+2}

this expression is expressed in terms of 1, 2 and 3. and hence we can make a circuit which consist only of 1 ohm, 2 ohms and 3 ohms and whose equivalent resistance is 11/3. that is :

1643115406019

3.20 (b)

(ii) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine the to get an equivalent resistance of (ii) \frac{11}{5}\Omega

Answer:

Connect 2 Ω and 3 Ω resistor in parallel and 1 Ω resistor in series to it

Equivalent Resistance R = {1/(1/2 + 1/3)} + 1 = 6/5 + 1

R = 11/5 Ω

3.20 (b)

(iii) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (iii) 6 Ω

Answer:

1 Ω+2 Ω+ 3 Ω= 6 Ω, so we will combine the resistance in series.

3.20 (b)

(iv) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them
to get an equivalent resistance of (iv) \frac{6}{11}\Omega

Answer:

Connect all three resistors in parallel.

Equivalent resistance is R = 1/(1/1 + 1/2 + 1/3) = (1x 2 x 3)/(6 + 3 + 2)

R = 6/11 Ω

3.20 (c)

(a) Determine the equivalent resistance of networks shown in
Fig. 3.31.

1643115440516

Answer:

It can be seen that in every small loop resistor 1 ohm is in series with another 1 ohm resistor and two 2 ohms are also in series and we have 4 loops,

so equivalent resistance of one loop is equal to the parallel combination of 2 ohms and 4 ohm that is

Equivalent\ R_{loop}=\frac{2*4}{2+4}=\frac{8}{6}=\frac{4}{3}

now we have 4 such loops in series so,

Total\ Equivalent\ R_{loop}=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=\frac{16}{3}

Hence equivalent resistance of the circuit is 16/3 ohm.

3.20 (c)

(b) Determine the equivalent resistance of networks shown in Fig. 3.31.

1643115474562

Answer:

It can be seen that all 5 resistors are in series, so

Equivalent Resistance = R + R + R + R + R = 5R

Hence equivalent resistance is 5R.

3.21 Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

1643257921855 Answer:

First, let us find the equivalent of the infinite network,

let equivalent resistance = R'

1643257970757

Here from the figure, We can consider the box as a resistance of R'

Now, we can write,

equivalent resistance = R'

' =[( R')Parallel with (1)] + 1 + 1

\frac{R'*1}{R'+1}+2=R'

R'+2R'+2=R'^2+R'

R'^2-2R'-2=0

R'=1+\sqrt{3},or1-\sqrt{3}

Since resistance can never be negative we accept

R'=1+\sqrt{3}

, We have calculated the equivalent resistance of infinite network,

Now

Total equivalent resistance = internal resistance of battery+ equivalent resistance of the infinite network

= 0.5+1+1.73

=3.23 ohm

V=IR

I=\frac{V}{R}=\frac{12}{3.23}=3.72A

Hence current drawn from the 12V battery is 3.72 Ampere.

3.22 (e) Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

1643258240599

(e) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

Answer:

No, the circuit would not work properly for very low order of Voltage because the balance points would be near point A and there will be more percentage error in measuring it. If we add series resistance with wire AB. It will increase the potential difference of wire AB which will lead to a decrease in percentage error.

3.23 Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

1643258276723 Answer:

Given,

the balance point of cell in open circuit = l_1=76.3cm

value of external resistance added = R=9.5\Omega

new balance point = l_2=64.8cm

let the internal resistance of the cell be r .

Now as we know, in a potentiometer,

r=\frac{l_1-l_2}{l_2}*R

r=\frac{76.3-64.8}{64.8}*9.5=1.68\Omega

hence the internal resistance of the cell will be 1.68 \Omega

Chapter 3 physics class 12 ncert solutions are essential for students preparing for board exams, JEE, and NEET. This chapter covers basic concepts like electric current and resistance. Understanding these concepts is important for board exams and helps in competitive exams like JEE and NEET. Class 12 current electricity ncert solutions also forms the foundation for advanced studies in physics and electrical engineering. So, it's a key chapter for success in exams and future studies.

Class 12 Physics NCERT Solutions for Current Electricity: Chapter- Wise

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Current electricity class 12 exercise solutions: Important Formulas and Diagrams

  • Current density

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1694164952162

Where: is the angle between the normal to the area and direction of current

Relation between current density and electric field

1694164953143

where = conductivity and = resistivity or specific resistance of the substance

  • Ohm’s Law

1694164951070

1694164952653

  • Resistivity

1694164951636

Where:

m is the mass, n is the number of electrons per unit volume, e is the charge of the electron and is the relaxation time

  • Grouping of Resistance

Series Grouping of resistance

1694165015527

Parallel Grouping of Resistance

1694165014961

Heat developed in a resistor

16941650160201694165014423

  • Kirchoff's Current law(KCL)

1694165013898

1694165013385

1694165012615

  • Kirchoff's voltage law(KVL)

1694165016529

1694165017660

1694165018309

  • Wheatstone's Bridge

1694165018817

1694165019903

1694165019365 (Balanced condition)

NCERT Class 12 Physics Chapter 3: Important Topics

Current Electricity Class 12 solutions for exercise questions are important from the board as well as competitive exam point of view. Following are the main topics covered in ch3 Physics Class 12.

  • Electric current- The concept of electric current, drift velocity, mobility etc are covered in the first part of the NCERT chapter on current electricity. All the formulas and concepts in the NCERT Class 12 Physics are important for the CBSE board exam and questions based on these parts are covered in the Current Electricity Class 12 NCERT solutions

  • Ohms law- Ohms law gives a linear relationship between voltage and current at a given temperature. And also discuss that not all conductors obey ohms law. There are ohmic and non-ohmic conductors. From here the term resistance is introduced and the questions related to this law is discussed in the NCERT Solutions for Class 12 Physics Chapter 3.

  • Resistance and Resistivity-The concepts of resistance and resistivity are discussed here. The variations of resistance and resistivity with temperature and colour code of resistance are discussed in chapter 3 Physics Class 12. Questions related to this are discussed in current electricity NCERT solutions.

  • Cells and their combinations-The concepts of cells, their series and parallel combinations are discussed in the CBSE Class 12 chapter 3 Current Electricity.

  • Kirchoff's law-This section discusses two important laws used to solve circuits namely Kirchoff's current law and voltage law. This part is important as far as competitive exams are concerned.

  • Wheatstone's bridge, meter bridge and potentiometer- The concepts of Wheatstone's bridge and meter bridge along with their balanced condition are discussed here. Meter bridge is an application of balanced Wheatstone's bridge. The concept of the potentiometer and its applications to compare EMFs of cells and also to find the internal resistance of cells are discussed in the chapter on current electricity. The questions from this part have high weightage for NEET and JEE main exams. 1 question from this part can be expected for JEE Main and NEET. To practice questions on this refer to chapter 3 Physics Class 12 NCERT Solutions.

Importance of Solutions of NCERT for Class 12 Chapter 3 Current Electricity in Board Exams:

As far as the CBSE board exam and competitive exams are considered, the solutions of NCERT Class 12 Physics chapter 3 are important. In NEET and JEE Main previous year papers, 7 to 10% questions are asked from this chapter. In CBSE board exam 2019, 6 marks questions are asked from current electricity. Use these CBSE NCERT solutions for Class 12 Physics chapter 3 PDF for a better score in board and competitive exams.

Benefits of NCERT Class 12 Physics Chapter 3 Solutions

  • This NCERT Solutions for Class 12 Physics chapter 3 will help in practising questions for annual exams.

  • Students can solve and match their answers from this Class 12 Physics Chapter 3 NCERT solutions.

  • These current electricity Class 12 NCERT solutions will give proper answers that will help students in exams to score good marks.

  • In case of any doubt, while solving any question given in chapter 3 Physics Class 12, students can refer to this solution to know how to solve the questions and what is the answer to a question.

Key Featuers of Class 12 Physics Chapter 3 Exercise Solutions

  1. Comprehensive Coverage: These class 12 current electricity ncert solutions cover all the topics and questions presented in Chapter 3, ensuring a thorough understanding of "Current Electricity."

  2. Detailed Explanations: Each solution provides step-by-step explanations, making complex concepts more accessible to students.

  3. Clarity and Simplicity: The current electricity class 12 exercise solutions are written in clear and simple language, aiding students in better comprehension.

  4. Practice Questions: Exercise questions are included, allowing students to practice and self-assess their understanding.

  5. Preparation for Exams: These class 12 physics chapter 3 ncert solutions help students prepare effectively for board exams, JEE, NEET, and other competitive exams.

  6. Foundation for Future Studies: The concepts covered in this chapter are essential for advanced topics in physics and electrical engineering.

  7. Interconnected Concepts: It establishes connections to later chapters, emphasizing its importance as a foundational chapter.

  8. Free Access: These solutions are available for free, making them easily accessible to all students.

These features make NCERT Class 12 Physics Chapter 3 Solutions a valuable resource for students to excel in their studies and exams.

Also Check NCERT Books and NCERT Syllabus here:

Subject wise solutions

Also, check

Excluded Content:

Certain portions of Chapter 3, "Current Electricity," have been omitted:

  • Section 3.7 about the resistivity of various materials, including Tables 3.1 and 3.2, as well as information about carbon resistors and their colour coding.
  • Section 3.10 discussing combinations of resistors in series and parallel.
  • Example 3.5.
  • Sections 3.15 on the meter bridge and 3.16 on the potentiometer.
  • Exercises 3.3, 3.4, 3.10, and the range of exercises from 3.14 to 3.23.

Frequently Asked Question (FAQs)

1. How important is the NCERT Solutions for Class 12 Physics chapter 3 for CBSE board exam?

In CBSE board exam, around 8 to 10% questions can be expected from the chapter Current Electricity. Certain papers of CBSE ask around 15% questions from NCERT chapter 3 Current Electricity. To score well in the exam follow NCERT syllabus and the exercise given in the NCERT Book. To practice more problems can refer to NCERT exemplar.

2. What are the important topics of Current Electricity?

If we analyse the previous year NEET, JEE Main and board papers we can see that the topics like meter bridge, potentiometer and problems using kcl and kvl and combinations of resistance and cells are frequently asked.

3. What is the weightage of current electricity in JEE main?

In JEE main, 2 to 3 questions can be expected from the chapter on Current Electricity. The topics like Potentiometer, meter bridge, KVL and KCL are important.

4. Whether NCERT is enough to cover Current Electricity for NEET?

The concepts in the current electricity chapter of NCERT Class 12 Physics are integral parts of NEET exam. Along with NCERT, practicing with previous year papers and mock tests would be enough.

5. What is the weightage of current electricity for NEET exam?

Current electricity carries 8% of weightage on an average for NEET exam.

6. How important is the NCERT Solutions for Class 12 Physics chapter 3 for higher studies in the field of engineering?

Current electricity is the basics of electrical and electronics engineering-related branches. In these branches, analysis and designs of the circuit are important and the basic laws studied in the Class 12 Physics Chapter 3 NCERT solutions help for the same.

7. What are the main topics covered in Current Electricity Class 12 NCERT chapter?

The main topics are-

  • Concept of current
  • Drift velocity and mobility
  • Ohm's Law
  • Resistance and resistivity
  • Combination of resistors
  • Cells and combination of cells
  • Kirchhoff Law
  • Whetstones bridge, meter bridge and potentiometers

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


Best CBSE schools in Delhi: Click Here


In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

Choosing Your Stream:

  • Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

  • Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

  • Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

  • Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
  • NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

Looking Ahead (2025 Admissions):

  • Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
  • NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

  • High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

  • Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

  • Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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